Problem: You have found the following ages (in years) of all 5 gorillas at your local zoo: $ 20,\enspace 22,\enspace 1,\enspace 7,\enspace 4$ What is the average age of the gorillas at your zoo? What is the standard deviation? You may round your answers to the nearest tenth.
Solution: Because we have data for all 5 gorillas at the zoo, we are able to calculate the population mean $({\mu})$ and population standard deviation $({\sigma})$ To find the population mean , add up the values of all $5$ ages and divide by $5$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{5}} x_i}{{5}} $ $ {\mu} = \dfrac{20 + 22 + 1 + 7 + 4}{{5}} = {10.8\text{ years old}} $ Find the squared deviations from the mean for each gorilla. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $20$ years $9.2$ years $84.64$ years $^2$ $22$ years $11.2$ years $125.44$ years $^2$ $1$ year $-9.8$ years $96.04$ years $^2$ $7$ years $-3.8$ years $14.44$ years $^2$ $4$ years $-6.8$ years $46.24$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{84.64} + {125.44} + {96.04} + {14.44} + {46.24}} {{5}} $ $ {\sigma^2} = \dfrac{{366.8}}{{5}} = {73.36\text{ years}^2} $ As you might guess from the notation, the population standard deviation $({\sigma})$ is found by taking the square root of the population variance $({\sigma^2})$ ${\sigma} = \sqrt{{\sigma^2}}$ $ {\sigma} = \sqrt{{73.36\text{ years}^2}} = {8.6\text{ years}} $ The average gorilla at the zoo is 10.8 years old. There is a standard deviation of 8.6 years.